1/5.8+1/8.11+...+1/x.(x+3)=1/6
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TN
4
19 tháng 8 2020
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{1}{6}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+3}=-\frac{3}{10}\)
\(\Leftrightarrow1\cdot10=-3\left(x+3\right)\)
\(\Leftrightarrow10=-3x-9\)
\(\Leftrightarrow10+9=-3x\)
\(\Leftrightarrow19=-3x\)
\(\Leftrightarrow x=-\frac{19}{3}\)
Đề sai à -.-
19 tháng 8 2020
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
=> \(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{1}{6}\)
=> \(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{6}:\frac{1}{3}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{1}{6}\cdot3=\frac{1}{2}\)
=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{1}{2}=-\frac{3}{10}\)
=> \(10=-3\left(x+3\right)\)
=> 10 = -9x - 9
=> 10 + 9x + 9 = 0
=> 19 + 9x = 0
=> 9x = -19
=> x = -19/9
BT
1
BD
6 tháng 5 2019
Đặt 1/5.8 + 1/8.11 +...+ 1 /x (x+3) = A
3A = 3/5.8 + 3/8.11 +...+ 3/x (x+3)
3A = 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/x - 1/x+3
3A = 1/5 - 1/x + 3
3A = ( 3+x)-5/5x +15
A =[ ( 3+ x ) - 5 / 5x + 15 ] : 3
A = x + ( - 2 ) / 5x + 15
Ta có :
A + 27/480
= x + ( - 2 ) / 5x + 15
=> x + ( - 2 ) = 27
=> 5x + 15 = 480
* Làm nốt *
#Louis
H
6 tháng 5 2019
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{27}{480}\)
\(=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{27}{480}\)
\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{27}{480}\)
\(=\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{27}{480}\)
\(=\frac{1}{5}-\frac{1}{x+3}=\frac{27}{480}.3\)
\(=\frac{1}{5}-\frac{1}{x+3}=\frac{81}{480}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{81}{480}=\frac{15}{480}=\frac{1}{32}\)
\(\Rightarrow x+3=32\)
\(\Rightarrow x=32-3=29\)
DK
0
LM
1
16 tháng 4 2021
Ta có: \(\dfrac{k}{x.\left(x+k\right)}=\dfrac{x+k-x}{x.\left(x+k\right)}=\dfrac{1}{x}-\dfrac{1}{x+k}\)
nên áp dụng ta có:
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\)
\(=\dfrac{1}{5}-\dfrac{1}{x+3}\)
Nên $\dfrac{1}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right)=\dfrac{1}{3}.(\dfrac{1}{5}-\dfrac{1}{x+3})$
Đến đây là làm được rồi nha
22 tháng 9 2016
a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+....+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(=3.\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=\frac{101}{1540}.3\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x.3}=\frac{303}{1540}\)
\(=\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(=\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(=\frac{1}{x+3}=\frac{1}{308}\)
\(x+3=308\)
\(\Rightarrow x=305\)
1/5.8+1/8.11+....+1/x(x+3)=1/6
=> 3/5.8 + 3/8.11 +.....+ 3/x(x+3)=1/2
=> 1/5-1/x+3=1/2
=> 1/x+3=1/5-1/2=-3/10=1/-10/3
=>x+3=-10/3=>x=-19/3
\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{1}{6}\)
\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\cdot\left(x+3\right)}=\frac{1}{2}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{x}-\frac{1}{x+3}=\frac{1}{2}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{1}{2}\)
\(\frac{1}{x+3}=\frac{-3}{10}\)
\(\Rightarrow\left(-3\right)\left(x+3\right)=10\)
\(\Rightarrow x+3=\frac{-10}{3}\)
\(\Rightarrow x=\frac{-19}{3}\)
Vậy \(x=\frac{-19}{3}\)